2.6.3 So­lu­tion herm-c

Ques­tion:

Show that the op­er­a­tor $\widehat 2$ is a Her­mit­ian op­er­a­tor, but $\widehat{\rm i}$ is not.

An­swer:

By de­f­i­n­i­tion, $\widehat 2$ cor­re­sponds to mul­ti­ply­ing by 2, so $\widehat 2g$ is sim­ply the func­tion $2g$. Now write the in­ner prod­uct $\left\langle\vphantom{2g}f\hspace{-\nulldelimiterspace}\hspace{.03em}\right.\!\left\vert\vphantom{f}2g\right\rangle$ and see whether it is the same as $\left\langle\vphantom{g}2f\hspace{-\nulldelimiterspace}\hspace{.03em}\right.\!\left\vert\vphantom{2f}g\right\rangle$ for any $f$ and $g$:

$$\langle f\vert\widehat 2 g\rangle = \int_{\mbox{\scriptsize ... ...ll }x} (2 f)^* g{ \rm d}x = \langle\widehat 2 f\vert g\rangle$$

since the com­plex con­ju­gate does not af­fect a real num­ber like 2. So $\widehat 2$ is in­deed Her­mit­ian.

On the other hand,

$$\langle f\vert\widehat{\rm i}g\rangle = \int_{\mbox{\scripts... ...m i}f)^* g{ \rm d}x = - \langle\widehat{\rm i}f\vert g\rangle$$

so $\widehat{\rm i}$ is not Her­mit­ian. An op­er­a­tor like $\widehat{\rm i}$ that flips over the sign of an in­ner prod­uct if it is moved to the other side is called skew-Her­mit­ian. An op­er­a­tor like $\widehat 2+\widehat{\rm i}$ is nei­ther Her­mit­ian nor skew-Her­mit­ian.